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shusha [124]
3 years ago
8

Adjectives for the word probable are......

Physics
2 answers:
iren [92.7K]3 years ago
5 0

the answer for the question is D. all of the above

Rina8888 [55]3 years ago
4 0
The answer is D all of the above.
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Vilka [71]

yes o=====/::::::::::::::::>

8 0
3 years ago
This type of wave is reflected by particles in the ionosphere
Alex787 [66]
I think the answer is <span>AM radio </span><span>waves. Im not sure but if its right can you mark my brainliest?

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4 0
4 years ago
If a 75-kg skater starts his skate at 8.0m, at his lowest point (height = 0), what is
Tanya [424]
Mgh= 1/2 m v^2
gh= 1/2 v^2 9.8 * 8=  1/2 v^2
solve for v
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6 0
3 years ago
Is everyone in your class able to hear a quiet sound equally well?
Sergio039 [100]

Answer:

No

Explanation:

Loudness describes how people perceive sound (see loudness). ... If people could hear equally well at all frequencies, the contour lines would be flat because the same measured sound intensity would be perceived to be equally loud regardless of the sound frequency. In fact, people do not hear as well at low frequencies.

8 0
3 years ago
A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th
sleet_krkn [62]

The electric potential V(z) on the z-axis is :  V = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z  

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq =  б( 2πR ) dr

dV  \frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }  ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = \int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,

 V = \pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]

     = \pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} }  -z ]

Therefore the electric potential V(z) = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

Also

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0
3 years ago
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